Factor the following expression: $x^2 - 16x + 60$
Solution: When we factor a polynomial, we are basically reversing this process of multiplying linear expressions together: $ \begin{eqnarray} (x + a)(x + b) &=& xx &+& xb + ax &+& ab \\ \\ &=& x^2 &+& {(a + b)}x &+& {ab} \end{eqnarray} $ $ \begin{eqnarray} \hphantom{(x + a)(x + b) }&\hphantom{=}&\hphantom{ xx }&\hphantom{+}&\hphantom{ (a + b)x }&\hphantom{+}& \\ &=& x^2 & & {-16}x& +& {60} \end{eqnarray} $ The coefficient on the $x$ term is $-16$ and the constant term is $60$ , so to reverse the steps above, we need to find two numbers that add up to $-16$ and multiply to $60$ You can try out different factors of $60$ to see if you can find two that satisfy both conditions. If you're stuck and can't think of any, you can also rewrite the conditions as a system of equations and try solving for $a$ and $b$ $ {a} + {b} = {-16}$ $ {a} \times {b} = {60}$ The two numbers $-10$ and $-6$ satisfy both conditions: $ {-10} + {-6} = {-16} $ $ {-10} \times {-6} = {60} $ So we can factor the expression as: $(x {-10})(x {-6})$